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(20 points) Given the ODE:
y
00
+ 4
y
0
+ 3
y
= 2 sin
x
(a)
Find the complementary solution.
Solution:
The characteristic equation is
r
2
+ 4
r
+ 3 = 0
which has roots
r
=

1 and
r
=

3. Therefore the complementary solution is
y
c
=
c
1
e

x
+
c
2
e

3
x
.
(b)
Find a particular solution.
Solution:
We note that there is no duplication between the function
f
(
x
) = 2 sin
x
and the complementary solution. We look for a solution
y
p
of the form
y
p
(
x
) =
A
cos
x
+
B
sin
x.
Then
y
0
p
(
x
) =

A
sin
x
+
B
cos
x
y
00
p
(
x
) =

A
cos
x

B
sin
x.
Substituting this into the ODE
y
00
p
+ 4
y
0
p
+ 3
y
p
= 2 sin
x
gives the equation

A
cos
x

B
sin
x
+ 4(

A
sin
x
+
B
cos
x
) + 3(
A
cos
x
+
B
sin
x
) = 2 sin
x
or
(2
A
+ 4
B
) cos
x
+ (2
B

4
A
) sin
x
= 2 sin
x.
Equating the coeﬃcients on the right and the left gives us the following two equations
for
A
and
B
.
2
A
+ 4
B
=
0

4
A
+ 2
B
=
2
which has the unique solution
A
=

2
/
5 and
B
= 1
/
5. The particular solution is
therefore
y
p
(
x
) =

2
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 Fall '07
 TERRELL,R
 Differential Equations, Equations

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